Thread: Equation of the tangent to a perpendicular lin

Can someone help me to solve this : Find the equation of the tangent to the curve x² -2xy + 2y² -7x + 6y + 6 = 0 which is perpendicular to 6x + 5y -4 = 0

Originally Posted by mgmanoj

Can someone help me to solve this : Find the equation of the tangent to the curve x² -2xy + 2y² -7x + 6y + 6 = 0 which is perpindendicular to 6x + 5y -4 = 0

Use implicit differentiation to find .

Find a point on the curve where .

After differentiating the equation I got
dy = 2x-2y-7
dx 2x-4y-6

the slope of the tangent perpendicuar to 6x + 5y-4 = 0 is 45/6 or 5/6 - I got confused here.

Originally Posted by mgmanoj

After differentiating the equation I got
dy = 2x-2y-7
dx 2x-4y-6

the slope of the tangent perpendicuar to 6x + 5y-4 = 0 is 45/6 or 5/6 - I got confused here.

You know Solve for either and substitute into the curve.

Originally Posted by mgmanoj

After differentiating the equation I got
dy = 2x-2y-7
dx 2x-4y-6

the slope of the tangent perpendicuar to 6x + 5y-4 = 0 is 45/6 or 5/6 - I got confused here.

The equation 6x+ 5y- 4= 0 is the same as y= -(6/5)x+ 4/6 and has slope -6/5. The "negative reciprocal" of that is 6/5 and is the slope of the perpendicular line. I don't know where you got "45/6"! Did you mean "4/6"? Remember that the slope of the line, the number multiplying x, determines the direction, not the "constant" term.